OxfordAQA International AS and A-level Physics: Vectors, equilibrium of forces and moments (9630)

This guide follows the OxfordAQA International AS and A-level Physics (9630) specification, Version 4.4, sections 3.2.1 (Scalars and vectors) and 3.2.2 (Moments). Both sections are shared International AS and A-level content, so they apply to everyone taking the qualification. Every worked example uses data given in the question; where gravitational field strength is needed, take g = 9.81 N kg⁻¹.

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What this unit covers

Section	What you must be able to do
3.2.1	Describe scalars and vectors, with examples: velocity and speed, mass, force and weight, acceleration, displacement and distance
3.2.1	Add vectors by calculation or by scale drawing
3.2.1	State the conditions under which two or three coplanar forces meeting at one point are in equilibrium
3.2.1	Appreciate that an object at rest, or moving at constant velocity, is in equilibrium
3.2.2	Define the moment of a force about a point as force × its perpendicular distance from the point
3.2.2	Define a couple and its moment; apply the principle of moments
3.2.2	Explain centre of mass; know that a uniform regular solid has its centre of mass at its centre

1. Scalars and vectors (3.2.1)

A scalar is fully described by its size. A vector needs a size and a direction.

Scalar	Vector counterpart
distance	displacement
speed	velocity
mass	weight (a force)
(no scalar partner)	acceleration, force

Distance is the total length of path travelled. Displacement is the shortest separation between start and finish, given with its direction. Speed is distance divided by time; velocity is displacement divided by time. Weight is the gravitational force on the mass, so it is a vector pointing towards the centre of the planet.

Worked example 1. A walker goes 45 m east, then 28 m north, in 50 s.

Distance = 45 + 28 = 73 m. Displacement = √(45² + 28²) = √2809 = 53.0 m. Direction: tan θ = 28/45, so θ = 31.9° north of east. Average speed = 73/50 = 1.46 m s⁻¹. Average velocity = 53.0/50 = 1.06 m s⁻¹, directed 31.9° north of east.

The velocity is smaller than the speed because the path is not straight.

2. Adding vectors (3.2.1)

The sum of vectors is called the resultant. Add vectors tip to tail: the resultant runs from the tail of the first to the tip of the last.

Right-angled vectors: Pythagoras and trigonometry

Worked example 2. A boat heads straight across a river at 3.0 m s⁻¹ relative to the water. The current is 1.2 m s⁻¹ downstream, at 90° to the heading.

Resultant speed = √(3.0² + 1.2²) = √10.44 = 3.2 m s⁻¹ (2 s.f.). Angle from the heading: tan θ = 1.2/3.0, so θ = 21.8° downstream of the heading.

Vectors at any angle: cosine rule or components

Two vectors a and b drawn tail to tail with angle θ between them have a resultant R where

R² = a² + b² + 2ab cos θ

Note the plus sign: drawn tip to tail, the interior angle of the triangle is 180° − θ, which is why the usual cosine rule has a minus. The components method avoids this confusion. Resolve each vector along two perpendicular axes, add the components on each axis, then recombine.

Worked example 3. Forces of 40 N and 55 N act at a point, with 70° between them. Find the resultant.

Take the 40 N force along the x-axis. Sum of x-components = 40 + 55 cos 70° = 40 + 18.81 = 58.81 N. Sum of y-components = 55 sin 70° = 51.68 N. R = √(58.81² + 51.68²) = 78.3 N. Angle from the 40 N force: tan φ = 51.68/58.81, so φ = 41.3°.

Check: R² = 40² + 55² + 2(40)(55) cos 70° = 6130, so R = 78.3 N. ✓

Resolving a single vector

A vector F at angle θ above the horizontal has horizontal component F cos θ and vertical component F sin θ.

Worked example 4. A force of 85 N acts at 35° above the horizontal.

Horizontal = 85 cos 35° = 69.6 N. Vertical = 85 sin 35° = 48.8 N.

Scale drawing

You can add vectors without trigonometry by drawing.

1. Choose a scale that fills the page, for example 1 cm to 10 N, and state it.
2. Draw the first vector to scale with a sharp pencil, with its direction measured by protractor from a stated reference line.
3. Draw the second vector from the tip of the first, at its correct angle.
4. Join the start to the end. Measure the length and convert using the scale, then measure the angle.

Quote the answer to the precision the drawing allows, usually 2 significant figures.

3. Equilibrium of coplanar forces at a point (3.2.1)

Coplanar forces lie in the same plane. An object is in equilibrium if the net force acting on it is zero. Then, by Newton’s first law, it is either at rest or moving with constant velocity. Zero resultant force does not mean zero velocity. A car cruising at a steady 20 m s⁻¹ along a straight road has its driving force equal to the drag, and is in equilibrium.

For two forces to be in equilibrium they must be equal in magnitude and opposite in direction, along the same line.

For three forces acting at a point, the conditions are:

1. The resultant of any two is equal and opposite to the third.
2. Drawn tip to tail, the three force vectors form a closed triangle.
3. The sum of the components along any chosen direction is zero.

Worked example 5. A lamp of mass 6.0 kg hangs at the junction of two cables. One cable (T₁) is 40° above the horizontal on the left; the other (T₂) is 65° above the horizontal on the right. Find both tensions.

Weight W = 6.0 × 9.81 = 58.86 N downwards.

Horizontal: T₁ cos 40° = T₂ cos 65°. Vertical: T₁ sin 40° + T₂ sin 65° = 58.86.

From the first, T₂ = T₁ (cos 40°/cos 65°) = 1.813 T₁. Substitute: T₁ (0.6428 + 1.813 × 0.9063) = 58.86, so 2.2856 T₁ = 58.86. T₁ = 25.8 N and T₂ = 46.7 N.

Check: horizontal components are 19.7 N each; vertical total is 16.6 + 42.3 = 58.9 N. ✓

Worked example 6. A smooth ramp is at 20° to the horizontal. A 15 kg block is held at rest by a rope parallel to the ramp. Find the tension and the normal contact force.

W = 15 × 9.81 = 147.2 N. Along the ramp: T = W sin 20° = 50.3 N. Perpendicular to the ramp: R = W cos 20° = 138 N.

Resolving along and perpendicular to the ramp gives one unknown per equation.

4. Moment of a force (3.2.2)

The moment of a force about a point is the force multiplied by the shortest distance from the point to the force’s line of action:

moment = F × d

The unit is N m. Moments are described as clockwise or anticlockwise.

If the force is not perpendicular to the line joining the point to the force’s point of application, use the perpendicular distance to the line of action. For a force F at angle θ to a rod of length L pivoted at one end, d = L sin θ.

Worked example 7. A force of 80 N pulls on a spanner of length 0.30 m, at 40° to the spanner. Find the moment about the nut.

d = 0.30 sin 40° = 0.193 m. Moment = 80 × 0.193 = 15.4 N m.

Pulling at 90° would give 24 N m.

5. Couples (3.2.2)

A couple is a pair of coplanar forces, equal in size and opposite in direction, acting along different lines. Their vector sum is zero, so a couple turns an object without pushing it anywhere. The moment of a couple is one force multiplied by the perpendicular distance between the two lines of action:

moment of couple = F × d

Worked example 8. A driver turns a steering wheel by pushing up with 14 N on one side and pulling down with 14 N on the opposite side, at opposite ends of a diameter of 0.36 m, both forces tangential.

Moment = 14 × 0.36 = 5.04 N m.

If the forces are not perpendicular to the line joining their points of application, use the perpendicular distance between the lines. Two 25 N forces acting 0.20 m apart along a rod, each at 60° to the rod, have separation d = 0.20 sin 60° = 0.173 m, so the moment is 25 × 0.173 = 4.33 N m.

6. The principle of moments (3.2.2)

Principle of moments: when a body is in equilibrium, the total clockwise moment about any chosen point equals the total anticlockwise moment about that same point.

Method for a beam problem:

1. Draw all forces, including weights acting at the centre of mass, and supports.
2. Choose a pivot through which an unknown force passes. That force then has zero moment and drops out.
3. Equate clockwise and anticlockwise moments.
4. Resolve forces vertically to find the remaining unknown.

Worked example 9. A uniform plank of length 5.0 m and mass 20 kg rests on supports P and Q, 0.5 m and 3.5 m from the left end. A child of mass 30 kg stands 4.0 m from the left end. Find the support forces.

Take moments about P. The plank’s weight acts at 2.5 m, 2.0 m from P. The child is 3.5 m from P. Q is 3.0 m from P.

Anticlockwise (Q): R_Q × 3.0. Clockwise: (20 × 9.81 × 2.0) + (30 × 9.81 × 3.5) = 392.4 + 1030.05 = 1422.45 N m. R_Q = 1422.45/3.0 = 474 N.

Vertical equilibrium: R_P + R_Q = 50 × 9.81 = 490.5 N, so R_P = 16.4 N.

P carries very little. P lifts off when the child is at x from the left end with 20 × 1.0 = 30 × (x − 3.5), so x = 4.17 m.

7. Centre of mass (3.2.2)

The centre of mass is the point through which the whole weight of a body can be considered to act. A body supported at its centre of mass balances. The centre of mass of a uniform regular solid is at its geometric centre, for example the middle of a uniform rod, cube or sphere. It need not lie inside the object: a uniform ring has its centre of mass at the centre of the hole.

Worked example 10. A uniform rod 1.2 m long has a mass of 0.90 kg. A 0.30 kg mass is fixed at one end. Find the balance point.

Let the balance point be d from the loaded end. The rod’s weight acts at 0.60 m from that end. Moments about the balance point: 0.30 × d = 0.90 × (0.60 − d). 0.30d = 0.54 − 0.90d, so 1.20d = 0.54. d = 0.45 m from the loaded end.

Common errors

• Using the rod’s full length as the arm of an angled force.
• Using sin for the vertical component when the angle is measured from the vertical. Draw the triangle first.
• Saying “equilibrium means not moving”. Constant velocity also counts.
• Taking moments about different points on the two sides of the equation.
• Putting a plank’s weight at an end instead of at its centre.
• Calling a couple a “resultant force”. Its resultant force is zero; its effect is a turning effect.

Official syllabus

OxfordAQA International AS and A-level Physics (9630), specification Version 4.4, sections 3.2.1 and 3.2.2, published by OxfordAQA.